How do you prove that the order of a in Zn (where n is an integer greater than or equal to 1) viewed as a group of order n with respect to addition is
n/(gcd(a,n))?
Let $\displaystyle G$ be a cyclic group (finite). And let $\displaystyle |G|=n$. It means the $\displaystyle G$ has a generator that is $\displaystyle <a>=G$ for some $\displaystyle a\in G$.
Let $\displaystyle b\in G$ thus, $\displaystyle b=a^c$ for $\displaystyle 1\leq c\leq n$.
We need to find the smallest $\displaystyle m$ such as,
$\displaystyle b^m=(a^c)^m=a^{cm}=e$
By the properties of cyclic groups and the fact that $\displaystyle a$ is a generator it is equivalent to saying $\displaystyle n$ divides $\displaystyle cm$.
Thus, we need the smallest $\displaystyle m$ such that,
$\displaystyle \frac{cm}{n}$ is an integer.
We can write it as, (by dividing by $\displaystyle d=\gcd (c,n)$
$\displaystyle \frac{m(c/d)}{(n/d)}$
But,
$\displaystyle \gcd \left( c/d,n/d \right)=1$
Thus,
$\displaystyle n/d$ divides $\displaystyle m$.
The smallest such $\displaystyle m$ is of course,
$\displaystyle n/d$
Thus, the order of any element is,
$\displaystyle \frac{n}{\gcd(c,n)}$